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CAT 2020 Slot 2 — QA Question 17

2 Variable EquationsEasy

Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

Answer & solution

  • A

    27

  • B

    36

  • 33

  • D

    30

Solution

Easy

Let the pencil cost be xx and sharpener cost x+2x+2. Equate the two spends (Aron vs Aditya) and simplify; the sharpener terms cancel, giving a clean relation x(p−10)=20x(p-10)=20. Both xx and pp are positive integers, so find the least valid pp, then total pencils is 3p3p.

1

Set up the spend equation. Aron: pp pencils, ss sharpeners. Aditya: 2p2p pencils, s−10s-10 sharpeners. Pencil price xx, sharpener price x+2x+2. Equal spends:

px+s(x+2)=2px+(s−10)(x+2)\begin{aligned} &px + s(x+2) = 2px + (s-10)(x+2) \end{aligned}
2

Simplify. Expand both sides; the sx+2ssx+2s terms cancel.

px=2px−10x−20⇒ 20=px−10x⇒ x(p−10)=20\begin{aligned} &px = 2px - 10x - 20\\ &\Rightarrow\ 20 = px - 10x\\ &\Rightarrow\ x(p-10) = 20 \end{aligned}
3

Minimize pp. We need p>10p>10 (so the bracket is positive) and p−10p-10 a positive divisor of 2020. The smallest such pp is p=11p=11 (with x=20x=20).

pmin⁡=11\begin{aligned} &p_{\min} = 11 \end{aligned}
4

Total pencils. Aron buys pp, Aditya buys 2p2p, together 3p3p.

3pmin⁡=3×11=33\begin{aligned} &3p_{\min} = 3\times 11 = 33 \end{aligned}
Minimum total pencils=33\text{Minimum total pencils} = 33
CAT 2020 Slot 2 QA Q17: Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as man — Solution | TheCATExam